Thursday, April 4, 2013
Browse »
home»
circuit
»
dissipation
»
l200
»
power
»
reducing
»
resistor
»
series
»
using
»
Reducing L200 Power Dissipation Circuit using Series Resistor
It’s good to reduce the power dissipated by the device. Using resistor connected in series to the input (the left figure) is a simple and economic method to reduce the device input-output differential voltage. This is the figure of the design circuit;
Here’s the formula for calculating R:
R= [Vi min- (Vo+Vdrop)]/Io
Vdrop = minimum differential voltage between the input and the output of the device at current Io
Vin min = minimum voltage
Vo = Output voltage
Io = output current
Resistor R can be connected between pins 1 and 2 of the IC instead of in series with the input if the load is constant (the right figure). So, part of the load current flows through the device and part through the resistor. This configuration is available when the minimum current by the load is:
Io min = Vdrop/R
Reducing L200 Power Dissipation Circuit using Series Resistor
It’s good to reduce the power dissipated by the device. Using resistor connected in series to the input (the left figure) is a simple and economic method to reduce the device input-output differential voltage. This is the figure of the design circuit;
Here’s the formula for calculating R:
R= [Vi min- (Vo+Vdrop)]/Io
Vdrop = minimum differential voltage between the input and the output of the device at current Io
Vin min = minimum voltage
Vo = Output voltage
Io = output current
Resistor R can be connected between pins 1 and 2 of the IC instead of in series with the input if the load is constant (the right figure). So, part of the load current flows through the device and part through the resistor. This configuration is available when the minimum current by the load is:
Io min = Vdrop/R
Subscribe to:
Post Comments (Atom)
No comments:
Post a Comment
Note: Only a member of this blog may post a comment.